How to prove that a finite-dimensional linear subspace is a closed set

Given a linear space $V$, a field $F$, a norm $||\cdot||$ on $V$ and a Base $B$.

How do I prove that the subspace span{$b_1,b_2,\ldots,b_n$} where $b_i \in B$ is a closed set under the topology that is created from the metric space that the norm creates?

Is it generally true? Do I need $V$ to be a Banach space? Or do I need $F$ to be the real numbers?

Solution 1:

You only need that the ground field $K$ is a complete normed field, e.g. $K \in \{\mathbb{R},\mathbb{C}\}$.

If $(x^{(k)})_k$ is a sequence in $U := \langle b_1, \dotsc, b_n \rangle$ which converges to some $x \in V$ then $(x^{(k)})_k$ is a Cauchy sequence. Because $K$ is a complete normed field the finite dimensional normed space $U$ is complete. Therefore $(x^{(k)})_k$ converges to some $y \in U$. By the uniqueness of limits in $V$ we already have $x = y \in U$. So $U$ is closed.

Notice that the statement does not necessarily hold for normed spaces over non-complete normed fields , even if all occuring vector spaces are finite-dimensional. Take for example the $\mathbb{Q}$-vector spaces $\mathbb{Q} \subseteq \mathbb{Q}[\sqrt{2}]$.